Draw all of the Lewis structures of NO2+ which obey the octet rule and use this Lewis structure or these resonance structures to predict how many covalent bonds connect each oxygen atom in the real structure to the central N atom. Assume that the octet rule is followed for the N atom when you draw y…
The Correct Answer and Explanation is :
Lewis Structures of ( \text{NO}_2^+ ):
The nitronium ion (( \text{NO}_2^+ )) consists of one nitrogen atom bonded to two oxygen atoms, with an overall +1 charge. Here’s the step-by-step procedure for drawing the Lewis structure(s):
- Count the total valence electrons:
- Nitrogen (( \text{N} )) has 5 valence electrons.
- Each oxygen (( \text{O} )) has 6 valence electrons, so for two oxygens, that’s ( 6 \times 2 = 12 ).
- Since the ion has a +1 charge, subtract 1 electron.
- Total = ( 5 + 12 – 1 = 16 ) valence electrons.
- Distribute electrons to satisfy the octet rule:
- The nitrogen atom is the central atom, as it is less electronegative.
- Form two ( \text{N=O} ) double bonds to satisfy the octet rule for nitrogen and the oxygens.
- This uses ( 4 \times 2 = 8 ) electrons in bonds, leaving 8 electrons to distribute as lone pairs on the oxygens.
- Assign formal charges:
- Formal charge = Valence electrons – Non-bonding electrons – (Bonding electrons / 2).
- In the structure with two ( \text{N=O} ) double bonds, the formal charge is 0 for each oxygen and +1 for nitrogen. This matches the overall charge.
Resonance Structures:
There is only one valid resonance structure for ( \text{NO}_2^+ ) because both ( \text{N=O} ) bonds are equivalent due to symmetry. No lone pair can form a resonance structure that maintains the octet rule.
Bond Prediction:
In the real structure, both ( \text{N=O} ) bonds are identical due to delocalization of electrons, with each being a bond order of 2.0 (double bond). This arises from the equivalence of the two ( \text{N=O} ) bonds in the resonance structure.
Explanation:
The Lewis structure illustrates how the octet rule is satisfied, with nitrogen forming two double bonds to oxygen. The symmetry and delocalization in the molecule imply identical bond lengths and strengths for both ( \text{N=O} ) bonds, making them true double bonds.